General solution of a differential equation

Dec 2008
105
3
Hi,

I have to solve the general equation for this differential

\(\displaystyle \frac{d^2y}{dx^2} - 6\frac{dy}{dx} + 9y = 50\sin(x)\)

I have the homogeneous equation as \(\displaystyle (A + Bx)e^{3x}\)

However now that I am trying to get the partial equation I am stuck, I am unsure as what to use for the \(\displaystyle y_{PI} = k(something)\) I have tried using sin(x) but that gets difficult to untangle

Thanks

Beard
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
You will need both sine and cosine:

y= A cos(x)+ B sin(x)

y'= -Asin(x)+ Bcos(x)

y"= -Acos(x)- Bsin(x)

y"- 6y+ 9y= -Acos(x)- Bsin(x)- 6Asin(x)+ 6Bcos(x)+ 9Acox(x)+ 9Bsin(x)

= (-A+ 6B+ 9A) cos(x)+ (-B- 6A+ 9B) sin(x)

= (8A+ 6B) cos(x) + (8B- 6A) sin(x)= 50 sin(x)

Solve 8A+ 6B= 0, 8B- 6A= 0 for A and B.
 
  • Like
Reactions: Beard
Dec 2008
105
3
You will need both sine and cosine:

y= A cos(x)+ B sin(x)

y'= -Asin(x)+ Bcos(x)

y"= -Acos(x)- Bsin(x)

y"- 6y+ 9y= -Acos(x)- Bsin(x)- 6Asin(x)+ 6Bcos(x)+ 9Acox(x)+ 9Bsin(x)

= (-A+ 6B+ 9A) cos(x)+ (-B- 6A+ 9B) sin(x)

= (8A+ 6B) cos(x) + (8B- 6A) sin(x)= 50 sin(x)

Solve 8A+ 6B= 0, 8B- 6A= 0 for A and B.
I wasn't actually expecting all of that, I was just asking for the first bit y= A cos(x)+ B sin(x) but thanks anyway.